# Prove that $e^{\sum 1/p_k^2} > \pi/2$

The $k$th prime is $p_k.$ Prove (if inclined, and without a machine) that

$$\exp\left({\sum_{k=1}^{\infty}\frac1{p_k^2}}\right) > \frac{\pi}{2}$$

• Where is this question from? Knowing the source can hint at the method of solution. My first guess at a solution would be to rewrite the left hand side of the equation as an infinite product and then to compare the terms of the product to the Wallis formula for $\pi/2$, perhaps using $e^x>1+x$, but I can't say for sure that this idea pans out. – Aaron Aug 4 '13 at 1:58
• @Aaron: I did the calculation and noticed I could not prove it. A quick search here did not turn it up. It seems familiar but I do not recall seeing it here. It's not a textbook sort of inequality. – daniel Aug 4 '13 at 2:31
• This can be verified by machine by summing the primes up to 250, but is there any reason to think that there should be a proof "by hand"? i.e., what makes this inequality different than putting $1.5717$ on the right-hand-side (which would also be a true inequality, and tighter)? – usul Aug 4 '13 at 3:01
• @Aaron's suggestion isn't quite good enough. It gives: $\exp\left( \sum_p \frac1{p^2}\right) = \prod_p \exp\left( \frac1{p^2}\right) > \prod_p \left( 1 + \frac1{p^2} \right) = \prod_p \left( \left(1 - \frac1{p^4}\right)\sum_n \frac1{p^{2n}}\right) = \left( \prod_p \sum_n \frac1{p^{2n}} \right) \prod_p \left(1 - \frac1{p^4} \right) = \frac{\pi^2}6 \prod_p \left( \sum_n \frac1{p^{4n}} \right)^{-1} = \frac{\pi^2}6 \frac{90}{\pi^4} = \frac{15}{\pi^2}$, which is too low by about 0.05. Maybe using $\exp(x) > (1+\frac x N)^N$ for all $N$? – Theo Johnson-Freyd Aug 4 '13 at 3:58
• @TheoJohnson-Freyd Using exact value of $\exp{\frac{1}{p^2}}$ for p=2,3,5 then we can fix the gap. – asatzhh Aug 4 '13 at 4:46

## 3 Answers

I thought I saw an answer here using the prime zeta function, which looked correct, so I didn't develop my answer. But as there are requests in the comments, here it is:

We start with Aaron's suggestion in the comments:

$$\exp\left( \sum_p \frac1{p^2}\right) = \prod_p \exp\left( \frac1{p^2}\right) > \prod_p \left( 1 + \frac1{p^2} \right) = \prod_p \left( \left(1 - \frac1{p^4}\right)\sum_n \frac1{p^{2n}}\right) = \left( \prod_p \sum_n \frac1{p^{2n}} \right) \prod_p \left(1 - \frac1{p^4} \right) = \frac{\pi^2}6 \prod_p \left( \sum_n \frac1{p^{4n}} \right)^{-1} = \frac{\pi^2}6 \frac{90}{\pi^4} = \frac{15}{\pi^2}$$

Throughout $p$ ranges over primes and $n$ ranges over nonnegative integers. I used the usual sum-to-product formula $\zeta(s) = \sum_n \frac1{n^s} = \prod_p \left( \sum_n \frac1{p^{ns}}\right)$, which follows from (and is equivalent to) unique factorization (and I think is due to Euler). I also used known zeta values $\zeta(2) = \frac{\pi^2}6$ and $\zeta(4) = \frac{\pi^4}{90}$. (I don't know a quick proof of the second of these; the first follows from two expansions of $\iint_0^1 \frac{\mathrm d x\, \mathrm d y}{1 + xy}$, if my memory is correct, and is in Proofs from the Book regardless.)

Unfortunately, estimating $\frac{15}{\pi^2}$ gives an answer that's a little less than $\frac{\pi}2$, but we'd win if we could make up a factor of $(\frac\pi 2 - \frac{15}{\pi^2}) / (\frac{15}{\pi^2}) < 3.36\%$. So let's see if we can squeeze a little bit more out. In the comments asatzhh suggests looking at explicit values for small $p$. We have: $$\frac{\exp(\frac14)}{1 + \frac14} > 1.0272,\ \frac{\exp(\frac19)}{1 + \frac19} > 1.0057, \ \frac{\exp(\frac1{25})}{1 + \frac1{25}} > 1.0007$$ which you can get, e.g. from Taylor expansion.

Thus our estimate above is an underestimate by at least $2.72\% + .57\% + .07\% = 3.38\%$. This completes the proof.

I should admit, I've been getting the decimal expansions from a calculator, and rounding in whichever direction makes the estimate worse. So I don't know how far out you need to look in Taylor expansion to get them by hand. But the Taylor expansion of $\exp(x)$ converges very quickly, and it's easy to estimate its error. And $\pi$ has many famous series expansions; see http://en.wikipedia.org/wiki/Approximations_of_%CF%80.

• The full infinite sum is known to great precision, oeis.org/A085548 . The closest thing from the ISC was 1000000/2211179, in any case about 0.45224742 isc.carma.newcastle.edu.au/standardCalc – Will Jagy Aug 4 '13 at 19:49
• (+1) for your answer. And it was my answer that used Prime Zeta Function and it wasn't correct, so I retracted it before it got any undeserved attention -_-; sheepish grin – TenaliRaman Aug 4 '13 at 22:20
• @TenaliRaman, Ah, sorry. Can I ask what was the error? Does the Mobius function change sign too often or something? – Theo Johnson-Freyd Aug 5 '13 at 3:59
• (+1) I'm happy to see that my idea could be salvaged. I'm curious if there is another solution that doesn't require getting an estimate and then fixing it. And more importantly, I'm curious if there is a solution which can be done without the aid of a calculator. – Aaron Aug 5 '13 at 4:31
• @TheoJohnson-Freyd I was bounding the series $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} \ln(\zeta(2k))$ below with just the first term in the series. The first term gives $\ln(\pi^2/6)$ which as observed in the comments is slightly larger the final sum and hence the bound as used clearly isn't true. Maybe there is a way to salvage it yet by considering few more terms from the series and showing that it does bound the series below, but my math-fu skill are lacking for that :-). – TenaliRaman Aug 5 '13 at 5:31

This answer is quite similar to TenaliRaman's and contains the details of my comments to the question. The basic idea is that the "prime zeta function" is comparable to the logarithm of the actual zeta function, via the Euler product:

$$\zeta(s) = \sum_{n > 0} \frac{1}{n^s} = \prod_{\text{p prime}} \frac{1}{1 - p^{-s}}.$$

If we take the logarithm, we get (with some necessary analysis to prove that the homomorphism property applies to the infinite product in this case):

$$\log \zeta(s) = \sum_p \log \frac{1}{1 - p^{-s}} = \sum_p \sum_{n > 0} \frac{(p^{-s})^n}{n}.$$

The last equality is just a well-known power series for $\log (1 - x)$:

$$\log (1 - x) = -\sum_{n > 0} \frac{x^n}{n}.$$

Anyway, taking just the first term of the $n$ sum in the previous equation for each $p$, we get

$$\log \zeta(s) > \sum_p p^{-s}.$$

Taking $s = 2$ gives the statement of the problem, and using the known value $\zeta(2) = \pi^2/6$, we have

$$\sum_p p^{-2} < \log \frac{\pi^2}{6} \iff \exp \sum_p \frac{1}{p^2} < \frac{\pi^2}{6}.$$

The right-hand side is greater than $\pi/2$, so this inequality is consistent with the problem, but without a proof that the difference between the two sides of the inequality above is less than $\pi^2/6 - \pi/2$, it doesn't actually finish it.

• By the way, my point in the comments was that the approximation $\exp \sum_p \frac1{p^2} \approx \frac{\pi^2}6$ is pretty close. You proved that it's $<$, whereas my comment gave $\exp \sum_p \frac1{p^2-1} > \frac{\pi^2}6$. In principal, this might let us complete this proof. Can we bound $\exp \sum_p \left( \frac1{p^2-1} - \frac1{p^2}\right) = \exp \sum_p \frac1{p^2(p^2-1)}$ to less than $\frac{\pi}6(\pi-3)$? Probably not... – Theo Johnson-Freyd Aug 7 '13 at 2:11

An early attempt which is clearly wrong. Making this into a community wiki, in case someone wants to take a stab at it and see if something can be salvaged here.

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The exponent is called as the Prime Zeta Function. From Mathworld , $$\sum_{k = 1}^{\infty} \frac{1}{p_k^2} = \sum_{k = 1}^{\infty} \frac{\mu(k)}{k} \ln(\zeta(2k))$$ where $\mu(n)$ is the Möbius function and $\zeta(n)$ is the Riemann zeta function. Therefore, $$\sum_{k = 1}^{\infty} \frac{1}{p_k^2} > \frac{\mu(1)}{1} \ln(\zeta(2)) = \ln\left( \frac{\pi^2}{6} \right)$$ $$\exp\left( \sum_{k = 1}^{\infty} \frac{1}{p_k^2} \right) > \frac{\pi^2}{6} > \frac{3\pi}{6} = \frac{\pi}{2}$$

• Your last inequality is false, and further more even if it was true, the mobius function changes sign infinitely often so you would still have to prove that the terms don't cancel out part of the first term (which they do). – Ethan Aug 4 '13 at 4:05
• Yes, realized that a while ago. Proving things without sleep is a bad idea. – TenaliRaman Aug 4 '13 at 4:08